C OMPOSITIO M ATHEMATICA
P ETER RUSSELL
Simple Galois extensions of two-dimensional
affine rational domains
Compositio Mathematica, tome 38, no 3 (1979), p. 253-276
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COMPOSITIO MATHEMATICA, Vol. 38, Fasc. 3, 1979, pag. 253-276
© 1979 Sijthoff &#x26; Noordhoff International Publishers-Alphen aan den
Printed in the Netherlands
Rijn
SIMPLE GALOIS EXTENSIONS OF
TWO-DIMENSIONAL AFFINE RATIONAL DOMAINS
Peter Russell
Introduction
field and B k[2]. (If R is a ring, we write R[n] for the
polynomial ring in n variables over R.) Let A C B, where A is an
affine factorial k-algebra. It was shown in [5] (under some mild
additional restrictions on A) that if B A[t] with t E qt(A) (the field
of quotients of A), then A k[2]. D. Wright [9] proved that this is true
also if t" E qt(A) with t lying in some extension field of qt(A) and
n &#x3E; 1. We in turn extend the ideas of [5] and [9] to show: Suppose
B A[t] is a simple extension of A and Galois over A in the sense
that G = AutA B is finite and qt(BG) qt(A). Then A k[2]. (The
restrictions on A mentioned above are again needed. See 2.4 for a
precise statement.) Let us note that the results of [5] and [9] as well as
[7] have been generalized in another direction in [6].
The proof of the above result breaks up naturally into two steps:
Let k be
a
=
=
=
=
=
=
(i) Classification of actions by a finite group G on B such that B is
a simple extension of BG. It turns out that G fixes a variable in
B and Bo - k[2]. This we show in section 1. (The argument is
quite brief if card G is prime to char k and somewhat involved
otherwise.)
(ii) Analysis of A C BG C B with qt(A) qt(BG). One finds A = k [21
by an argument very close to the one used to settle the
birational case in [5]. This is done in section 2.
=
[9] one derives, in a by now familiar way,
concerning the cancellation problem for k[2] and the
problem of embedded planes from our main result. This is the content
As in [5], [6], [7] and
consequences
of section 3.
Using techniques from [6]
we
prove the
0010-437X/79/03/0253-24 $00.20/0
253
following results in section
254
4: Let K be a locally factorial Krull domain and F E B K[2] such
that for each prime P C K the canonical image of F in
Kp OK BIP(Kp @KB) (KP/PKpi2] is a variable. Then F is a variable
in B. As a consequence of this and the results of section 1 we can
settle a further special case of cancellation for k[2]: Let k be a locally
f actorial Krull domain and A a k-algebra such that A[1] = A[T] =
k[X, Y, Z] = k[3] with Z E A[Tn] for some rt &#x3E; 1 invertible in k. Then
=
=
A = k[2].
.
As a matter of notation, by a statement A = K[n] we always mean
that K is (in an obvious way) a subring of A and A is K-isomorphic
with K[n].
1 would like to acknowledge the hospitality and lively mathematical
environment of Bhaskaracharya Pratishtana of Poona, India, which I
enjoyed during the preparation of this paper.
1.
Let k be a commutative
subgroup. Put
ring,
B
a
k-algebra and
B is a simple extension of A, say B
b
E
B
can be written
any
Suppose
and
we
deduce the
The next point is
described will arise
=
G C Autk B
a
finite
A[t] with t E B. Then
f ollowing basic fact:
an
immediate consequence of (1.1). The situation
times in the sequel.
numerous
(1.2) Let K C B be
a
subring
such that B
=
K[t’] for some t’ E B.
255
REMARK: If R is a ring, we denote by R * the group of units and
R+
the additive group of R.
by
In 1.5 and 1.6 we collect some more or less well known facts about
additive polynomials. The proofs are included for the convenience of
the reader.
a
(1.5) Let K be a domain of characteristic exponent p and H C K+
finite subgroup. Put
Then
f H ( T ) is
a
monic p-polynomial in T, that is
with ai E K and p" card H. In particular, fH(T) is additive i.e., if R is
any K-algebra and Ti, T2 E R, then
=
PROOF: We may assume p &#x3E; 1. Let h E H. Then 1-lo:gi:5p-l (T - ih) =
TP - ThP-’. This proves the result in case card H p. If card H &#x3E; p,
we can write H
Hl (D H2 with card Hi card H, i 1, 2.
=
=
One finds
=
256
with H’ = f H2(H,). B y induction
nomials. Hence, so is fH.
on
card H,
f H, and f H2
(1.6) LEMMA: Let K be a field of characteristic
finite subgroup and
homomorphism. Then there exists
f(T) E K[T] such that
a
card H". Induction on card H
Uniqueness is obvious.
now
a
p &#x3E;
are
0, H
p-poly-
C
K+
a
unique additive polynomial
establishes the existence of f.
(1.7.1) LEMMA: Let K be a domain, B K[1] and G C AutK B a
finite subgroup. Then there exists a finite cyclic subgroup W C K*, a
finite subgroup H C K+ stable under multiplication by elements of W
and T E B with B K [ T ] such that
=
=
is a homomorphism and W ir(G) is a finite, hence cyclic, subgroup
of K*. If W {1}, set T X. Otherwise, choose e E G such that
w = at/1 generates W and put T = X + (w - l)-’b,,, Then e(T) = wT
and clearly G is the semi-direct product of .p) W and Ker 1T = H
E Ker ?r} C K+ with W operating on H by multiplication.
=
=
=
=
{acp l ’P
=
257
Let q card H. Since W operates
have q - 1 - 0 mod r and Wq= w for any
PROOF:
we
=
on
H
by multiplication,
Also,
w E W.
Be. Now B K[T] is integral of degree rq over both K[v]
and B G. Hence B G. CR K[v] n K[T] where K is the integral closure
of K, and it is easy to see that R K[v] since v is monic in T (see also
lemma 2.6.1 of [6]).
Let k be a field, p char k and B = k[2] for the remainder of this
section. We fix (for the moment) a system (x, y ) of variables for B. If
GA2, we write
cp E Autk B
Hence v E
=
=
=
=
=
to mean that cp
f 2 E B.
is the
k-homomorphism with
Put
and
(1.8) Let G C GA2 be a finite subgroup. As is well known (see [4],
theorem 3.3, for instance), GA2 is the amalgamated product of its
subgroups AF2 and E2. It follows that G is conjugate to a subgroup of
AF2 or E2 (see [8]) or, which is the same, is a subgroup of AF2 or E2
after suitable choice of (x, y).
(1.9)
LEMMA: Let B
with a, b E k. Then G
that B k[xi, YI] and
card k.
=
=
converse.
suppose each cp E G is
of the form
fixes a variable in B, i.e., there exist xl, YI such
CP(XI) Xl for all cp E G, if and only if card G:5
PROOF:
Clearly
If, say, G fixes y, then
prove the
[x, y]and
=
258
Then b is determined by a whenever cp = (x + a, y + b) E Gand we
can find an additive polynomial f 2(x) such that f2(a) = b as ç ranges
subgroups with
H stable under
multiplication by
elements of W.
259
with ai e k and p n
elements of W,
In
=
card H. Since H is stable under
multiplication by
particular,
Put
u
=
f(x, y) and
v
=
y". Then BG’
steps follow from 1.7.2 modulo
Making
use
of (2) and (3)
we
some
find
Hence,
The constant term w.r.t. T of 0 is
and the linear term is
=
k[u, y] and BG
=
k[u, v]. (Both
confusion in the notation.) The
260
B, and
we
have reached
a
contradiction.
PROOF: The "if" part of the theorem is obvious and we proceed to
prove the "only if" part. By 1.8 we may assume G C E2 or G C AF2.
Let cp E G and write
Then
implies order cp Y-’ char k
Hence
=
p.
261
1 for
follows from (6) and (7) that
w,y = 1 and a, 0 for all Ji E G. Hence G fixes y.
(1.2) Assume Vcp 1 for all ç E G. Suppose q = (x + h(y), y + a) E L
with a# 0. We may then assume a = 1. Note that L = Gc by (4).
Suppose am y m appears in h(y) with m + 14 0 mod p. Replacing x by
x - (am/m + l)ym+l we eliminate the y’"-term from h(y) without disturbing terms of higher degree or changing L. It is clear, therefore,
that we may assume
(I.1) Assume v,:,x-’
some cp
E G. It
=
=
where only terms bjyj with j4 0
17P 1 and we must have h1(yP)
=
Let
and
mod p
=
appear in
0, that is
h2(y). Since q
E
G,,
262
By 1.7.2, h E k[v] where v fH(Y) is an additive polynomial. We have
e(v) = fH(Y + a) = v + b with b = fH(a):;é 0 by (8). Replacing x by
x + g( v) with g( v ) E k [v] we do not alter the representation of ele0 and hence g(fH(Y + a’)) g(v) f or a’ E H.
ments of G’ c since fH(a’)
Repeating the argument given above we may assume, therefore, that
h 0. Proceeding this way we obtain L Gc G’C.
=
=
=
=
=
This discussion
(10)
can
be summarized
as
=
follows:
suitable choice of variables L is of one of the
types.
L1: There exists q = (x, y + 1) e L and every ip E L is of the
form .p = (x + h.", y + at/1) with a.", ht/1 E k.
L2: a, 0 for all Ji E L, that is L fixes y.
After
a
following
two
=
(1.2.1) Suppose G L. If L is of type L1, then G Gc by (4) and G
is isomorphic to a subgroup of k+ by 1.3. Hence G fixes a variable by
1.9. If L is of type L2, then G fixes y.
=
=
(1.2.2) Suppose
cyclic, subgroup of k* and
W is
r
=
a
finite, hence
card W &#x3E; 1. Let
It follows from (12) that we can determine cj such that (1- wl)cj +
b; 0 for all j. Replacing x by x’ (this does not change the type of L)
we may assume, therefore, that cp
(x, wy). Then L {1} by 1.10 and
x.
G fixes
=
=
=
263
Let H be the subgroup of k+ generated
It follows from the above that ;&#x3E;
and since
no
term
bmym with
m ---
by
0 mod
r
appears in
f,
Arguing as in (a) we may assume, after replacing x by x’ = x + gi(v)
with suitable gl( V) E k[ v], that hl( v) = h ’(y) = o. Again, this substitution does not change the type of L. In fact, if tp E L, then E H,
Since now h is an additive polynomial we can next change x to
x + g2(Y), where g2(Y) is additive, with the eff ect of making h
vanish. Again this does not affect the type of L, for now 4i(x’) =
x’
=
264
cp(t) - t divides both Il (a, - 1)x + b,y + ci and 12
a2x+(b2-1)y+ c2 by 1.1. Hence 11 and 12 are linearly dependent.
We conclude that for each cp E G one eigenvalue of Mcp is 1. Then
the other is det Mcp E k*, and one finds by a straightforward calculation that all Mcp can be simultaneously brought into lower triangular form by a linear change of variables. This brings us back to case
the other hand,
=
=
I.
(1.11.1) COROLLARY: Let the assumptions be as in 1.11 and let
fixed by G. Then there exists YI E B such that
B
and
G
GIG2 is a semi-direct product, where
k[Xh YI]
xl E B be a variable
=
=
with W and H subgroups of k* and
H by multiplication. Moreover,
k[xl]’ respectively
and W
acting
on
degree card H.
PROOF: This follows from 1.11, 1.7.1 and 1.7.2.
( 1.11.2) REMARK: If card G is prime to char k then G G2 = W is
cyclic. It is clear that the proof of 1.11 simplifies considerably under
=
this
assumption.
then ç
=
(x + h(y), y)
w.r.t. suitable
variables (x, y) for k[2].
In fact,
by 1.8 we may assume cp E AF2 or cp E E2. In the latter case
(x h(y), y + b ) with h(y) E k[y] and b E k. In the first case this
ç
form can be achieved by a linear change of variables. If b 0 0 then h
can be made to vanish by changing x to x’
x + g(y) with suitable
E
seen
in
1.2
we
have
of
the
as
g(y) k[y]
proof of 1.11.
=
+
=
REMARK: It would be highly desirable to find a proof of 1.11 that
does not make explicit use of the structure theorem for Autk k[2]
hidden in 1.8, particularly with a view of extending the result to
polynomial rings in three (or more) variables. If the latter were
265
possible,
one would have established, in
the cancellation property for k[2].
conjunction with 4.3 below,
2.
(2.1) LEMMA: Let A and A’ be noetherian domains with A C A’
such that qt(A) qt(A’). Suppose A is normal pref actorial, every
height 1 prime of A’ contracts to a hight 1 prime of A and A* A’*.
Then A A’.
=
=
=
Q and Q’ be the set of height 1 primes of A and A’
If
P E Q, then P is the radical of fA for some f E A
respectively.
A
is
(since
prefactorial) and f é A* A’*. If P’ is a minimal, hence
height 1, prime of fA’, then P’ fl A D fA and P’ fl A E Q by assumption. Hence P’ nA = P. Now Apl:J Ap and qt(Ap’) qt(A). Hence
A’, = Ap since A is normal. It follows that A’ c np’EQ’ Ap’ C
npEO Ap = A.
For the remainder of this section we assume
PROOF: Let
=
=
a field, A a finitely generated k-domain such that qt(A)
and in addition one of the following holds:
(2.2) k is
qt(kl’])
=
A is factorial and contains a field generator, i.e., there exists
f E A such that qt(A) k(f, g) for some g E qt(A);
(ii) char k 0 and A is factorial;
(iii) k is perfect and A is regular prefactorial.
(i)
=
=
(2.3) PROPOSITION: Let A be as in 2.2. Suppose
ACB’CB
where B
every
=
k[2],
height
1
B’ =
k[2],qt(A) qt(A’),
prime of B
=
contracts to
B
a
A[t] for some t E B, and
height 1 prime of B’. Then
=
A = k[2].
PROOF: We may
set of
assume
A C B’. Let S
=
{P1, ..., PJ be the (finite)
height 1 primes Pi of B’ such that A n P, mi is maximal. Note
S 0 0 by 2.1. Write Pi= fiB’ with fi E B’ and let Pi be a minimal
prime of f;B. Then Pin A = Mi, MiB c fiB C Pi and BIMIB
=
=
266
where t is the image of t mod M,B. It follows that
Now Pi n B’ is
Pi and we have
a
height
1
prime containing f;B’ = Pi, hence
It follows from (5) that B/Pi is integral
only one valuation at infinity). Hence
over
B’/P; (since B/Pi has
if M C B’ is a maximal ideal and fi E M, then there is
maximal ideal N C B such that N fl B’ = M.
(6)
Making use of (3), (4)
proof of 1.3 of [5] to find
and (7) we now proceed exactly
x, v E B’ such that
as
a
in the
We claim:
If aB’ gB’, then
qt(A)
=
fi is a factor of a in B’ and a E Mi (recall that
qt(B’)). Suppose this is the case. Arguing as in (6) we find a
some
267
maximal ideal N of B’ such that g E N and N n A Mi. But then
fi El N, and this is impossible since (/,,g)B’=BB It follows that
aB’ = gB’. Hence g E A and x is integral over A, so x E A since A is
normal.
Suppose (i) or (ii) of 2.2 holds. In view of (8) and (9) we find, as in
the proof of 1.3 in [5], that A k[x, uv] where u E k[x] is of minimal
degree such that uv E A.
Suppose (iii) of 2.2 holds. Using theorem 3.1 of [5] we find u E k[x]
such that A C k [x, uv ] A’ and every height 1 prime of A’ contracts
to a height 1 prime of A. Then A
A’ by 2.1.
We record the following more precise version of 2.3 established
during the proof.
=
=
=
=
x E A such that A = k[x][l], B’ =
A
B’
an irreducible f E k[x] such
and
either
or
there
exists
k[x][l]
that fB’ fl A is maximal and B/fB k[xllf"’. Moreover, if x has this
property and B’ = k[x, v], then A k[x, uv] for some u E k[x].
(2.3.1) COROLLARY: There exists
=
=
=
REMARK: Let the notation be as in 2.3.1. It seems highly likely that
B k[x][l] as well. We will show this under special circumstances in
the next section.
=
(2.4) THEOREM: Let A be as in 2.2. Assume
(i) A C B k[2] such that B A[t] for some t E A,
(ii) if G AutA B and B’ = BG, then qt(A) qt(B’).
=
=
=
=
Then
More
precisely, variables for A
PROOF : B is a
Also, B is integral
can
be chosen
as
in 2.3.1.
simple extension of B’ = BG. By 1.11.1, B’ k[2].
over BG. Hence all assumptions of 2.3 are satisfied.
=
3.
a k-algebra and FE A[T] = AU], Fe A. Put B =
The
canonical map from A to B is injective. We
A[T]/FA[T].
A
identif y with its image and write B A[t], where t is the image of
T.
Let A be
=
268
(3 .1 ) DEFINITION.
(i)
F is
(a)
a
Galois equation
F is
over
A
if
prime;
(3.2) THEOREM:
Let k be
a
perfect field and A
Suppose Z is a Galois equation
over
a
A. Then A =
k-algebra such that
k[2].
A is finitely generated over k, regular and factorial,
so 2.2(iii) holds. Also B = A[T]/ZA[T] = k[2]. Hence the theorem
follows from 2.4.
PROOF:
Clearly
A more precise description of F from which 3.3 will follow is given
in 3.8.4. We start by collecting some preliminary results. The first, a
substitute of sorts in positive characteristic for the epimorphism
theorem of Abhyankar and Moh [2], was obtained recently by R.
Ganong [3].
(3.4) PROPOSITION: Let k be a field and x E k[xi, y,] k[2] such that
k[Xh ylllx k[l]. Then there is a unique k(x)-valuation V of k(xl, YI)
not containing k(x)[xl,
K(x)[x1, YI],
y1], and the residue field L of V is purely
inseparable over k(x). Moreover, k[Xh YI] k[xl[’] if and only if L
k(x).
=
=
=
and either
=
269
or
PROOF: Consider
By 3.4 the valuation at infinity VI of qt(A’) (given by - degy) has a
unique extension V to qt(B’). Let Li and L be the residue fields of Vi
and V respectively and f [L:Li]. Then f p m for some m by 3.4.
Moreover, [qt(B’) : qt(A’)] ef, where e is the ramification index of V
1.
over VI. By 3.4 it is enough to show f
If (i) holds, then [qt(B’) : qt(A’)] r is prime to p, hence f 1. So
assume (ii) holds. Then G has a normal subgroup Gi of order p. Let cp
generate G1. By 1.11.3 we can assume that x,, YI have been chosen so
that W = (xi, y, + h(xl» with h(Xl) e k[x1]. Then B G. - k[2]. Moreover,
BGllx = k[l] by the argument used to establish (7) in the proof of 2.3.
By induction on card G, BGI k[x][’] and it is enough to prove the
lemma in case G Gi. Let - denote images mod x and write B/xB
k[t]. Then p(t) - tEk and Ç(gi) - gi = h(ii). If Ç # 1, then h(xl) E k
by 1.2, and Ç 1 implies h(il) = 0. Hence x1 h(xl) - c with c E k. If
h(XI)-C-:;éO, this implies x=axl+b with a E k * and b E k and we
are done. Otherwise h(xl)
c E k* and B’is unramified over A’. Then
V is ramified over V¡, that is e &#x3E; 1, and since ef p we have f
1.
The next result was mentioned without proof in [5], 3.6. We need it
now and briefly indicate a proof.
=
=
=
=
=
=
=
=
=
=
=
=
(3.6)
LEMMA: Let k be a
such that
field, Î
an
=
algebraic closure of k and
FE k[T, S] = k[2]
CÀ = {F = À 1 À
PROOF: Let A be the pencil of curves
E k} and
Ps the base points of ll (they are all at infinity). Then, as is
pi,
well known, it follows from (ii) that all members of A, including the
generic member, "go through each other" in the sense that the
...
,
multiplicity of (the proper transform of) C, at pi is independent of À.
Moreover, the generic member is a rational curve over k(F). On the
270
other hand, by (i) all base points of A are rational over k (they are on
Co which has one place at infinity rational over k), so the generic
member in fact is a rational curve over k(F), and this implies
k[S, T] k[F][l], as is again well known.
=
min(deg f, deg g).
the proof of 3.3. Clearly B is a simple extension of
BG:) A and we can choose variables xi, yi for B as in 1.11.1. We use
the notation established there and let p char k.
We
now
turn to
=
(3.8.1) There exists
PROOF:
(i) If
A
=
x
e A such that A =
BG,
our
k[X][1]
and B =
k[x][1].
claim follows from 1.11.1.
(ii) Suppose A C BG. Clearly 2.2 (i) holds for A k[2] and hence the
assumptions of 2.4 are satisfied. We choose x as in 2.3.1. Let k be an
algebraic closure of k and put B, = k Q9kB. Clearly if fi is an irreducible factor of f over k, then fi = ax + 8 with a E k * and 8 E k,
B IG t---e k[fl][l] and Btlfl = trI]. Noting Brl = j{£2] we find Brl/fl = k-111 by
the argument for (7) in the proof of 2.3. Since [qt(Brl): qt(BG)] =
card G2 = r with (r, p) = 1, we have B = k[fl][1] k[x ][1] by 3.5(i).
Similarly Bi = k[x ][1] now follows from 3.5(ii).
=
=
(a ) If G is of
constant
and B
assumption
(3) Suppose G is
=
type, then k is separable over k by
k[x][1] by [5], lemma
not of constant
1.5.
type. Then either there exists
271
REMARK: If char k 0, we can finish the proof of 3.3 by referring
to [6], theorem 2.6.2. The remainder of the proof is devoted mainly to
establishing that F is a variable in k(x) Q9k[x] A[T] in case char k
p &#x3E; 0. Even in characteristic 0, though, we have the bonus of finding
an explicit form for F.
=
=
PROOF: This is clear from 1.11.1 and 2.3.1.
or
Hence = acp = a." and it follows that a, is
constant
for .p E G.
Hence
there exist d EE k * and cE k[x, v] such that y
=
d(t + c).
Assume W = {1} and there exist hI, h2 EH such that hiO- 0,
i 1, 2 and hi ùÉ kh2. Given any such pair put .pi = (x, y + hi).
=
272
Then
and hence at/11 at/11t/12 al/12’ Again a, is independent of e and
(2) holds.
Suppose W {1} and H H’h with fixed h E k[x] and H’ C k+.
=
=
(y)
=
=
Moreover, if
1T
is
an
irreducible factor of u’, then
273
f (T) is monic of degree p" in T, u’pnf (c’/u’) = c’P" + u’a with
k[x, w]. So clearly U2U is a polynomial Û of minimal degree in
k[x] such that ûF’ E k[x, w, T] and it follows that F F". Let 1T be
Since
a
E
=
irreducible factor of U2. Then F reduces to
this is not possible since
an
UlC,rpn == 0 mod 1T,
and
Hence u2 = 1. Now let Tr be an irreducible factor of u’. Then F =
UtC,rpn - w mod 1T, and since (1) again holds, F is congruent to a
polynomial of degree 1 in w mod ir.
(ii) Let u’c c’ as in (i) and put upnl/u,pn = u,/u2 with u,, u2 E k [x] and
GCD(ul, U2) 1. One sees as before that U2 1 and F upnlF’ =
=
=
=
=
Then F reduces to UtC,pn - an, W pn’ --- 0 mod -r where
leading coefficient of g. Since p" &#x3E; p nt &#x3E; 1 this is again
incompatible with (1), so u 1 as claimed.
factor of
anl is the
u.
=
PROOF:
the
way that FTT’ the canonical image of F in (k[x]/17’)
is
[w, T], a variable if 7T E k[x] is irreducible and w 9 u. If, on
the other hand,
this follows from the considerations in
3.8.4(i). The conclusion now follows from theorem 4.1 below
same
ir( u,
274
(3.8.6) REMARK: Clearly if F is
as in (3.8.4) and f = fH for some
stable under multiplication by elements of W =
then F is a Galois equation over k[x, w] with group
G isomorphic to the semidirect product HW.
Let
us
note that if M?"
1, then the examples of "embedded planes"
all of the type described in [6], (3.8.2). Let us
obtained this way are
point out also that if F is
general is not Galois.
a
Galois
equation,
then F +
c
with
c
E k* in
4.
(4.1) THEOREM: Let K be a locally factorial Krull domain and
F E K[X, Y] K[2]. For each prime P C K let Lp qt(K/P), denote
by Fp the canonical image of F in Lp[X, Y] and assume Lp[X, Y] =
Lp[Fp][1]. Then K[X, Y] K[F][1].
=
=
=
((2) and (3) are immediate consequences of LP [X, Y] = Lp[Fp][I].)
Also, since F is a variable, hence transcendental, modulo each
maximal ideal of K, the content of F - F(O) is (1) and
Hence
generated
rank
1
projective
K-module.
Hence K
(4.2) REMARK: Let F (Fi,..., Fr) and X (X,, ..., Xr). Otherwise
keep the notation of 4.1, and assume Fi,..., F, are part of a system of
=
=
275
variables in Lp[X, Y] for each prime P C K. (1), (2) and (3) in the
above proof then hold. Moreover (4) is easy to establish if P (0)
since K is normal. Clearly (4) is trivial for P:;é 0 in case K is a
principal ideal domain. Hence the conclusion of the theorm holds in
that case.
=
(4.3) THEOREM: Let k be a locally factorial Krull domain and
k-algebra such that
Suppose Z E A[T"] with
n &#x3E;
A
a
1 and invertible in k. Then
PROOF: Suppose first that k contains a primitive n-th root of unity
w. Then Z E A[ T"] if and only if Z is fixed by the A-automorphism cp
We claim:
however, is impossible since cp(T)
Hence Tp 0 0.
If k does not contain
a
=
wT-:;é T whereas
primitive n-th
root of
unity,
cp(ab)
we
=
ab.
adjoin
one,
276
again calling it
w. Then k[w] D k is an integral extension, and so is
K.
If
P
is a prime of K, let P’ be a prime of K [w]such that
K[w]:)
P’ fl K P. Then Lp, D LP is a separable extension. By what we have
shown, Tp E Lp[X, Y] is a variable in Lp,[X, Y], and by [5], lemma
1.5, Tp is a variable in Lp[X, Y]. It follows from 4.1 that B
k[Z, U, T] for some U E B. Hence A B/TB = k[2].
=
=
=
REFERENCES
[1] S.S. ABHYANKAR, P. EAKIN and W. HEINZER: On the uniqueness of the
coefficient ring in a polynomial ring. J. Algebra 23 (1972) 310-342.
[2] S.S. ABHYANKAR and T.-T. MOH: "Embeddings of the line in the plane". J. Reine
Angew. Math. 276 (1975), 148-166.
[3] R. GANONG: On plane curves with one place at infinity. Thesis. McGill University,
1978.
[4] M. NAGATA: "On automorphism group of k[X, Y]". Lectures in Mathematics 5,
Kyoto University, Kinokuniya Bookstore, Tokyo.
[5] P. RUSSELL: "Simple birational extensions of two-dimensional affine rational
domains". Compositio Math. 33 (1976), 197-208.
[6] P. RUSSELL and A. SATHAYE: "On finding and cancelling variables in k[X, Y, Z]".
(To appear in J. Algebra).
[7] A. SATHAYE: On linear planes. Proc. Amer. Math. Soc. 56 (1976), 1-7.
[8] J.-P. Serre, Arbres, amalgames et SL2. Collège de France, 1968/69.
[9] D. WRIGHT: Cancellation of variables of the form bTn - a. (To appear).
(Oblatum 15-VIII-1977)
Department of Mathematics
McGill University
Montreal Quebec
Canada